Problem 20 - Calculus for Biology and Medicine

Question

Apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\left(3 x^{3}-3\right)\left(2-2 x^{2}\right), \text { at } x=0 $$

Step-by-Step Solution

Answer

The tangent line at \( x = 0 \) is \( y = -6 \).

Step 1: Identify Functions for Product Rule

Given the function \( f(x) = (3x^3 - 3)(2 - 2x^2) \), identify \( u = 3x^3 - 3 \) and \( v = 2 - 2x^2 \) as the functions to apply the product rule.

Step 2: Differentiate Each Function

Find the derivatives of \( u \) and \( v \):- For \( u = 3x^3 - 3 \), \( u' = 9x^2 \).- For \( v = 2 - 2x^2 \), \( v' = -4x \).

Step 3: Apply the Product Rule

Use the product rule for differentiation: \( f'(x) = u'v + uv' \).- Substitute \( u' = 9x^2 \), \( v = 2 - 2x^2 \), \( u = 3x^3 - 3 \), \( v' = -4x \).- Calculate: \( f'(x) = (9x^2)(2 - 2x^2) + (3x^3 - 3)(-4x) \).

Step 4: Simplify the Derivative

Simplify the expression for \( f'(x) \):- \( 9x^2(2 - 2x^2) = 18x^2 - 18x^4 \).- \( (3x^3 - 3)(-4x) = -12x^4 + 12x \).- Combine terms: \( f'(x) = 18x^2 - 18x^4 - 12x^4 + 12x = -30x^4 + 18x^2 + 12x \).

Step 5: Evaluate the Derivative at the Point

Plug \( x = 0 \) into the simplified derivative: - \( f'(0) = -30(0)^4 + 18(0)^2 + 12(0) = 0 \).

Step 6: Find the Tangent Line Equation

Evaluate the original function at \( x = 0 \):- \( f(0) = (3(0)^3 - 3)(2 - 2(0)^2) = -6 \).- The slope \( m = f'(0) = 0 \) and point \((0,-6)\).- Equation in slope-intercept form is \( y = mx + b = 0 \cdot x - 6 = -6 \).

Key Concepts

Product Rule

The product rule is an essential tool in calculus, especially when dealing with functions involving products of two or more terms. It's like having a special formula to differentiate these types of functions. The product rule states that if you have two functions multiplied together, say \( u(x) \) and \( v(x) \), the derivative of their product \( y = u(x) v(x) \) is calculated as follows: \[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \] This means you differentiate \( u \) while keeping \( v \) the same, plus differentiate \( v \) while keeping \( u \) unchanged. When applying the product rule:

• Start by identifying the functions \( u \) and \( v \) that make up the product.
• Find the individual derivatives \( u' \) and \( v' \).
• Plug these into the product rule formula.

For example, in our exercise, \( f(x) = (3x^3 - 3)(2 - 2x^2) \), we identified \( u = 3x^3 - 3 \) and \( v = 2 - 2x^2 \). After differentiating and applying the product rule, we found the derivative of \( f(x) \). Understanding the product rule helps solve many calculus problems and is a key concept in differentiation.

Differentiation

Differentiation is a core concept in calculus, allowing us to understand how a function behaves as its input changes. Essentially, it's about finding the "instantaneous rate of change" of a function, much like how speed measures how a car's position changes over time. The process involves taking a function and computing its derivative. This derivative gives us critical insights:

• It can tell us the slope of the function at any given point.
• Helps in identifying maxima, minima, and points of inflection.

In our example, differentiating the function \( f(x) = (3x^3 - 3)(2 - 2x^2) \) involves applying the product rule to get \[ f'(x) = -30x^4 + 18x^2 + 12x \]. Each term in the derivative corresponds to part of the rate at which \( f(x) \) changes with \( x \).Differentiation breaks down complex relationships in biological systems, analyzing rates like population growth or response to stimuli.Biologists use it to model and predict changes in systems based on variable shifts.

Tangent Line

The tangent line to a curve at a given point is like the curve's best linear approximation at that specific spot. In simpler terms, it's the straight line that "just touches" the curve at one point. This line has the same slope as the curve does at that point. Finding a tangent line involves:

• Calculating the derivative of the function to find the slope \( m \) at the given point.
• Determining the function value at the point to know where the line touches the curve.
• Using the point-slope formula to write the tangent's equation.

In slope-intercept form, the equation of a line is \( y = mx + b \). For our function, at \( x = 0 \), we found the slope \( m \) to be \( 0 \) and the point it touches \( f(0) = -6 \). Thus, the tangent line equation became \( y = -6 \), a horizontal line reflecting the flatness at that point. Tangent lines are instrumental in fields like biology for examining patterns and predicting how functions behave near specific values.